Question

For reaction \(A \rightarrow P\), rate constant \(k = 1.5 \times 10^3\ s^{-1}\) at \(27^\circ C\). If activation energy for the above reaction is \(60\ kJ\ mol^{-1}\), then the temperature (in \(^{\circ}C\)) at which rate constant \(k = 4.5 \times 10^3\ s^{-1}\) is ______. (Nearest integer) \[ \text{Given: } \log 2 = 0.30,\ \log 3 = 0.48,\ R = 8.3\ J\ K^{-1}\ mol^{-1},\ \ln 10 = 2.3 \]

Answer 1

Correct Answer: 41

Step 1: Understanding the Concept:
The variation of the rate constant with temperature is given by the Arrhenius Equation. We use the logarithmic form to solve for the unknown temperature.
Step 2: Key Formula or Approach:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
Note: \(2.303 \times R \approx 2.3 \times 8.3 = 19.09 \).
Step 3: Detailed Explanation:
Given:
\(k_1 = 1.5 \times 10^3 \text{ s}^{-1}\)
\(k_2 = 4.5 \times 10^3 \text{ s}^{-1} \implies \frac{k_2}{k_1} = 3\)
\(T_1 = 27 + 273 = 300 \text{ K}\)
\(E_a = 60 \text{ kJ/mol} = 60,000 \text{ J/mol}\)
Substitute into the equation:
\[ \log 3 = \frac{60000}{2.3 \times 8.3} \left( \frac{T_2 - 300}{300 T_2} \right) \]
\[ 0.48 = \frac{60000}{19.09} \left( \frac{T_2 - 300}{300 T_2} \right) \]
\[ 0.48 = 3143 \left( \frac{T_2 - 300}{300 T_2} \right) \]
\[ \frac{0.48}{3143} = \frac{T_2 - 300}{300 T_2} \implies 0.0001527 = \frac{1}{300} - \frac{1}{T_2} \]
\[ \frac{1}{T_2} = \frac{1}{300} - 0.0001527 = 0.0033333 - 0.0001527 = 0.0031806 \]
\[ T_2 = \frac{1}{0.0031806} \approx 314.4 \text{ K} \]
Convert Kelvin to Celsius:
\[ T_2 = 314.4 - 273 = 41.4^{\circ}C \]
Rounding to the nearest integer, we get 41.
Step 4: Final Answer:
The temperature is 41\(^{\circ}C\).